[QUOTE=science_man_88;475314]Two word come to mind [URL="https://en.m.wikipedia.org/wiki/Pigeonhole_principle"]pigeonhole priciple[/URL][/QUOTE]

please elaborate how the pigeonhole principle may be applied in order to prove the conjecture ...
i am not able to see that clearly - maybe i am blind ? :cry:

That's a nice one. Thanks, guptadeva!

For the benefit of the people who have trouble with inequalities:
When confused, try to go from abstract to concrete.

Example: why is the highlighted error above an error?
Here is why: consider a concrete similar argument.
Predicates: a >= 7, b >= 6.
Does it follow that a-b >= 1? Of course not (e.g. take a=b=10; take a=10, b=100, etc)

Here is how the correct reasoning could flow:
Predicates: a >= 7, b [B]<=[/B] 6.
Conclusion: a-b >= 1. This argument is correct.

[QUOTE=robtaylor501;475325]... I'm not ready just yet to change my definitions or notation ...
[/QUOTE]

there is really no need doing that !
i really like the style and presentation on your blog.
the translation into some mathematical formalism was just a necessity for me to better understand your concepts and ideas ...

[QUOTE=guptadeva;475329]please elaborate how the pigeonhole principle may be applied in order to prove the conjecture ...
i am not able to see that clearly - maybe i am blind ? :cry:[/QUOTE]

Divisibility on any arithmetic progression is governed by it. Take the first r entries in any arithmetic progression, either r divides into one of them, or it divide into none of the values in the progression. It can then be generalized to be univariate polynomial remainder theorem. Even Euler's generalized version of Fermat's theorem can be deduced from it. Okay, it's general remainder.

[QUOTE=guptadeva;475332]there is really no need doing that !
i really like the style and presentation on your blog.
the translation into some mathematical formalism was just a necessity for me to better understand your concepts and ideas ...[/QUOTE]

Thank you! I enjoy the blog. It's a good way to at least get ideas out into the world. 10metreh's counterexample still has me floored. Now I'm wondering if there is any way to calculate a lower bound on the number of primes needed when accommodating sets. I guess the question isn't as valuable, as 10metreh shows perfect accommodations with less than pi(x)+2 primes exist, thus throwing my inequalities out the window and also throwing the takeaways on Legendre's, etc. out the window. Moreover, (what I thought was) a nice philosophical takeaway goes out the window, as if that "shape" did in fact perfectly accommodate all set sizes and we let x tend to infinity, then the end result would be the layout of the set of integers, meaning all numbers were arranged in a perfect accommodation. It had a nice zen feel to it. It made sense. Oh well. C'est la vie.

These prime numbers! So unruly.

[QUOTE=robtaylor501;475343]Thank you! I enjoy the blog. It's a good way to at least get ideas out into the world. 10metreh's counterexample still has me floored. Now I'm wondering if there is any way to calculate a lower bound on the number of primes needed when accommodating sets. I guess the question isn't as valuable, as 10metreh shows perfect accommodations with less than pi(x)+2 primes exist, thus throwing my inequalities out the window and also throwing the takeaways on Legendre's, etc. out the window. Moreover, (what I thought was) a nice philosophical takeaway goes out the window, as if that "shape" did in fact perfectly accommodate all set sizes and we let x tend to infinity, then the end result would be the layout of the set of integers, meaning all numbers were arranged in a perfect accommodation. It had a nice zen feel to it. It made sense. Oh well. C'est la vie.

These prime numbers! So unruly.[/QUOTE]

I think it goes back to post 6. That's because unlike part of what I stupidly just said, the pigeonhole principle, doesn't gaurantee that same remainder will be 0. I think you should use:

[TEX]\phi(n) + \Omega(n)[/TEX] where phi is Euler's totient function, and Omega is prime factors counted once each.

[QUOTE=robtaylor501;475343]These prime numbers! So unruly.[/QUOTE]

yet there is order in the chaos ... search and you will find ... :bounce wave:

[QUOTE=robtaylor501;475343]Thank you! I enjoy the blog. It's a good way to at least get ideas out into the world. 10metreh's counterexample still has me floored. Now I'm wondering if there is any way to calculate a lower bound on the number of primes needed when accommodating sets. I guess the question isn't as valuable, as 10metreh shows perfect accommodations with less than pi(x)+2 primes exist, thus throwing my inequalities out the window and also throwing the takeaways on Legendre's, etc. out the window. Moreover, (what I thought was) a nice philosophical takeaway goes out the window, as if that "shape" did in fact perfectly accommodate all set sizes and we let x tend to infinity, then the end result would be the layout of the set of integers, meaning all numbers were arranged in a perfect accommodation. It had a nice zen feel to it. It made sense. Oh well. C'est la vie.

These prime numbers! So unruly.[/QUOTE]

ok ... i will give you some hopefully fruitful hints:

in order to find an expression or inequality for your 'accomodation lemma' you really do not need to consider perfect accomodations for all different arrangements.
it is sufficient to consider perfect accomodations of the actual arrangements for/of the sets of all odd numbers between n^2 and (n+1)^2 for increasing n ...
if you start to see a pattern in these arrangements you would then maybe be able to find a form of the general 'shape' (we really need a better word for that) and then try to apply induction from there

you could also consider taking one step back and include the prime 2 and all even numbers back into your considerations :exclaim:

another approach could be to succesively sieve all numbers which can be accomodated by the primes 2,3,5,... out from [n^2, (n+1)^2] and see if you might be able to start to see some pattern in the sets remaining ...

alternatively you could also start with a proof of the oppermann conjecture instead :smile:

finally you could also step out of the box and attempt to find some new relation or pattern among the prime numbers themself :innocent:

[QUOTE=science_man_88;475344]I think it goes back to post 6. That's because unlike part of what I stupidly just said, the pigeonhole principle, doesn't gaurantee that same remainder will be 0. I think you should use:

[TEX]\phi(n) + \Omega(n)[/TEX] where phi is Euler's totient function, and Omega is prime factors counted once each.[/QUOTE]

honestly thanks for your suggestion ...

your mind seems to be as twisted or convoluted as the prime numbers are spread among the natural numbers :razz:

[QUOTE=guptadeva;475380]honestly thanks for your suggestion ...

your mind seems to be as twisted or convoluted as the prime numbers are spread among the natural numbers :razz:[/QUOTE]

Well the prime numbers above 3 are in one of 2 arithmetic progressions with constant difference 6.

[QUOTE=science_man_88;475390]Well the prime numbers above 3 are in one of 2 arithmetic progressions with constant difference 6.[/QUOTE]

well ... so far so good
do you happen to know some necessary and sufficient condition for a number to be a prime - other than a sieve ?
a [B]simple[/B] unconditional deterministic primality test would be fine :smile: